M is a nxn matrix where all entries are 1's.
My conjecture is that n - 1 eigenvalues are 0 and the nth eigenvalues is one.
If this is correct, how do I go about proving it?
Using the steps:
$\displaystyle (i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n$
we obtain
$\displaystyle \displaystyle\begin{aligned}\begin{vmatrix} 1-\lambda & 1 & \ldots & 1\\ 1 &1-\lambda & \ldots & 1 \\ \vdots&&&\vdots \\ 1 & 1 &\ldots & 1-\lambda\end{vmatrix}&=\begin{vmatrix} 1-\lambda & \;\;1 & \ldots & \;\;1\\ \lambda &-\lambda & \ldots & \;\;0 \\ \vdots&&&\vdots \\ \lambda & \;\;0 &\ldots & -\lambda\end{vmatrix}\\&=\begin{vmatrix} n-\lambda & 1 & \ldots & 1\\ 0 &-\lambda & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & -\lambda\end{vmatrix}\\&=(n-\lambda)(-\lambda)^{n-1}\end{aligned}$
So, the eigenvalues are $\displaystyle \lambda=0$ (n-1-uple) and $\displaystyle \lambda=n$ (1-uple) if $\displaystyle n\geq 2$ . If $\displaystyle n=1$ we have only the eigenvalue $\displaystyle \lambda=1$ (1-uple).
In general with the steps
$\displaystyle (i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n$
we triangularize the determinant
$\displaystyle \begin{vmatrix} a & b & \ldots & b\\ b & a & \ldots & b \\ \vdots&&&\vdots \\ b & b &\ldots & a\end{vmatrix}=\ldots=[a+(n-1)b](a-b)^{n-1}$
first he subtracted the first row, from every other row. this changed all the 1's in the first column (except for the top entry) to -λ.
it also changed all the 1's in every other column to 0, except along the diagonal, where 1-λ becomes -λ.
then he sucessively added (it doesn't matter if you do it one at a time, or all at once) columns 2 through n, to the first column.
neither the row operation, nor the column operation changes the determinant, since no row or column was multiplied.
Now the eigenvectors.
For $\displaystyle \lambda =0$, I obtained
$\displaystyle \left\{\begin{bmatrix}-1\\n\\0\\ \vdots\\0\end{bmatrix},\begin{bmatrix}-1\\0\\n\\ \vdots\\0\end{bmatrix},\cdots,\begin{bmatrix}-1\\0\\0\\ \vdots\\n\end{bmatrix}\right\}$
However, I am finding it hard to solve for lambda = n
Is it just
$\displaystyle \begin{bmatrix}1\\0\\0\\ \vdots\\0\end{bmatrix}$
Shouldn't we get this:
$\displaystyle \begin{vmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{vmatrix}$
$\displaystyle x_2=-x_3-x_4-\cdots -x_n$
$\displaystyle x_i=0 \ \ i=2,...,n$
$\displaystyle x_1$ is a free variable.
So shouldn't it just be the n dimensional $\displaystyle e_1$ vector?